L'Hospital's rule

This may or not employ L'Hospital's, but here it is:

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.

Try using Maclaurin series if you know them.

Originally Posted by vince

u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true)

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.